1

我正在尝试为@NamedNativeQuery具有a 值的@ConstructorResult字段的类创建 a 。Setenum

兽医JPA.java：

@Entity
@Table(name = "veterinarians")
@Setter
@Getter
@NoArgsConstructor
@NamedNativeQueries({
        @NamedNativeQuery(
                name = VeterinarianJPA.FIND_ALL_VETS,
                query = "SELECT v.id, v.name, vs.specialisations " +
                        "FROM veterinarians v " +
                        "JOIN veterinarian_specialisations vs ON v.id = vs.vet_id",
                resultSetMapping = VeterinarianJPA.VETERINARIAN_RESULT_MAPPER
        )})
@SqlResultSetMappings({
        @SqlResultSetMapping(
                name = VeterinarianJPA.VETERINARIAN_RESULT_MAPPER,
                classes = @ConstructorResult(
                        targetClass = Veterinarian.class,
                        columns = {
                                @ColumnResult(name = "id", type = Long.class),
                                @ColumnResult(name = "name"),
                                @ColumnResult(name = "specialisations", type = Set.class)
                        }
                )
        )})
class VeterinarianJPA {

    static final String FIND_ALL_VETS = "net.kemitix.naolo.gateway.data.jpa.findAllVets";
    static final String VETERINARIAN_RESULT_MAPPER = "net.kemitix.naolo.gateway.data.jpa.Veterinarian";

    @Id
    @GeneratedValue
    private Long id;

    private String name;

    @ElementCollection
    @Enumerated(EnumType.STRING)
    @CollectionTable(
            name = "veterinarian_specialisations",
            joinColumns = @JoinColumn(name = "vet_id")
    )
    private final Set<VetSpecialisation> specialisations = new HashSet<>();
}

兽医.java：

public final class Veterinarian {

    private Long id;
    private String name;
    private Set<VetSpecialisation> specialisations;

    public Veterinarian() {
    }

    public Veterinarian(final long id,
                        final String name,
                        final Set<VetSpecialisation> specialisations) {
        this.id = id;
        this.name = name;
        this.specialisations = new HashSet<>(specialisations);
    }

    public long getId() {
        return id;
    }

    public String getName() {
        return name;
    }

    public Set<VetSpecialisation> getSpecialisations() {
        return new HashSet<>(specialisations);
    }

}

VetSpecialisation.java：

public enum VetSpecialisation {

    RADIOLOGY,
    DENTISTRY,
    SURGERY

}

当我尝试执行命名查询时：

entityManager.createNamedQuery(VeterinarianJPA.FIND_ALL_VETS, Veterinarian.class)
             .getResultStream()

我得到以下异常：

java.lang.IllegalArgumentException: Could not locate appropriate constructor on class : net.kemitix.naolo.entities.Veterinarian

    at org.hibernate.loader.custom.ConstructorResultColumnProcessor.resolveConstructor(ConstructorResultColumnProcessor.java:92)
    at org.hibernate.loader.custom.ConstructorResultColumnProcessor.performDiscovery(ConstructorResultColumnProcessor.java:45)
    at org.hibernate.loader.custom.CustomLoader.autoDiscoverTypes(CustomLoader.java:494)
    at org.hibernate.loader.Loader.processResultSet(Loader.java:2213)
    at org.hibernate.loader.Loader.getResultSet(Loader.java:2169)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1930)
    at org.hibernate.loader.Loader.executeQueryStatement(Loader.java:1892)
    at org.hibernate.loader.Loader.scroll(Loader.java:2765)
    at org.hibernate.loader.custom.CustomLoader.scroll(CustomLoader.java:383)
    at org.hibernate.internal.SessionImpl.scrollCustomQuery(SessionImpl.java:2198)
    at org.hibernate.internal.AbstractSharedSessionContract.scroll(AbstractSharedSessionContract.java:1058)
    at org.hibernate.query.internal.NativeQueryImpl.doScroll(NativeQueryImpl.java:217)
    at org.hibernate.query.internal.AbstractProducedQuery.scroll(AbstractProducedQuery.java:1462)
    at org.hibernate.query.internal.AbstractProducedQuery.stream(AbstractProducedQuery.java:1486)
    at org.hibernate.query.Query.getResultStream(Query.java:1110)

我希望 SQL 为多值Set而不是单个值返回多行，这导致构造函数不匹配。如何更改 SQL 以向构造函数生成正确的输入，或者是否需要进行其他配置更改？

4

1 回答 1

0

好吧，我不确定这是否可能以您想要的方式进行。但是您可以使用表格上的LISTAGG函数通过使用某种分隔符来内联。specialisationsspecialisationsveterinarians

所以查询应该是这样的：

SELECT v.id, v.name 
(SELECT LISTAGG(vs.type, ';') 
  WITHIN GROUP (ORDER BY vs.type) 
  FROM veterinarian_specialisations vs 
  WHERE vs.vet_id = v.id) specialisations
FROM veterinarians v;

该查询将返回 veterinarian 和他的分号分隔的专业：

1   NAME   DENTISTRY;RADIOLOGY

然后在您的Veterinarian类构造函数中，您必须将 String 结果重新映射回 Set of VetSpecialisation。我使用 Java 8 流 API 只是为了方便。

public final class Veterinarian {

private Long id;
private String name;
private Set<VetSpecialisation> specialisations;

public Veterinarian() {
}

public Veterinarian(final long id,
                    final String name,
                    final String specialisations) {
    this.id = id;
    this.name = name;
    this.specialisations = Arrays.asList(specialisations.split(";"))
        .stream()
        .map(VetSpecialisation::valueOf) //Map string to VetSpecialisation enum.
        .collect(Collectors.toSet());
}
于 2018-08-09T07:31:05.140 回答